RI sec 3 1997 EOY P2 Q15
(a) Solve [(2x + 3) / (x² - 1)] - 2 = 1 / (x + 1)
(b) A man finds that it takes 10 minutes less to cut his lawn with a new mower which cuts at 12.5 m² per minute faster than his old one. The area of his lawn is 320 m².
(i) Taking x to be the rate, in m² per minute, at which the new mower cuts, write down an expression, in terms of x, for the time taken to cut the lawn with the new mower.
(ii) Hence form an equation in x and show that it reduces to 2x² - 25x - 800 = 0
(iii) Solve this equation and find the time taken to cut the lawn with the new mower.
(iv) If his new mower consumes 1 litre of petrol per 10 minutes and each litre of petrol costs $1.25, calculate the amount of petrol used and the cost of the petrol.
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Answer:
(a) [(2x + 3) / (x² - 1)] - 2 = 1 / (x + 1)
(2x + 3) - 2(x² - 1) = (x² - 1) / (x + 1)
(2x + 3) - 2(x² - 1) = (x -1) ==> x² - 1 = (x + 1)(x - 1)
2x + 3 - 2x² + 2 = x - 1
2x² - x - 6 = 0
(2x + 3)(x - 2) = 0
x = -1.5 or x = 2
(b)
(i) Time taken = 320/x mins
(ii) 320/(x + 12.5) = 320/x + 10
-320/x + 320/(x + 12.5) = 10
[320x + 4000 - 320x] / [320/(x² - 12.5x)] = 10
400 = x² - 12.5x
x² - 12.5x - 400 = 0
2x² - 25x - 800 = 0 (shown)
(iii) 2x² - 25x - 800 = 0
Using formula, x = (-b ± √(b² - 4ac) ) / 2a,
x ≈ 27.2 or -14.7 (reject because x must be positive)
Hence, time taken = 320/27.2 = 11.76 mins
(iv) Amount of petrol used = 1/10 * 11.76 * $1.25 = $1.47
